If p(x)=x^{2}-4 x+3, evaluate : p(2)-p(-1)+pl | vedclass

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Question

We have $p(x)=x^{2}-4 x+3$

$	herefore p(2)-p(-1)+p\left(\frac{1}{2}ight)=\left(2^{2}-4 	imes 2+3ight)-\left\{(-1)^{2}-4(-1)+3ight\}+\left\{

Vedclass · Accepted Answer

$\frac{-31}{4}$

Vedclass · Answer

We have $p(x)=x^{2}-4 x+3$

$	herefore p(2)-p(-1)+p\left(\frac{1}{2}ight)=\left(2^{2}-4 	imes 2+3ight)-\left\{(-1)^{2}-4(-1)+3ight\}+\left\{\left(\frac{1}{2}ight)^{2}-4 	imes \frac{1}{2}+3ight\}$

$=(4-8+3)-(1+4+3)+\left(\frac{1}{4}-2+3ight)$

$=-1-8+\frac{5}{4}$

$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$

If $p(x)=x^{2}-4 x+3,$ evaluate $: p(2)-p(-1)+p\left(\frac{1}{2}\right)$

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